3.255 \(\int \frac {\sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=122 \[ \frac {2 a (7 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d \sqrt {\sec (c+d x)}} \]

[Out]

2/15*a*(7*A+15*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/5*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)
/d/sec(d*x+c)^(3/2)+2/15*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {4087, 4013, 3804} \[ \frac {2 a (7 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(2*a*(7*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Sqrt[a + a*Sec[c + d
*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d
*x]])

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a A}{2}+\frac {1}{2} a (2 A+5 C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {1}{15} (7 A+15 C) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a (7 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 68, normalized size = 0.56 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} (8 A \cos (c+d x)+3 A \cos (2 (c+d x))+19 A+30 C)}{15 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

((19*A + 30*C + 8*A*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d*Sq
rt[Sec[c + d*x]])

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fricas [A]  time = 0.78, size = 86, normalized size = 0.70 \[ \frac {2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + {\left (8 \, A + 15 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + (8*A + 15*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sqrt(a*sec(d*x + c) + a)/sec(d*x + c)^(5/2), x)

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maple [A]  time = 2.81, size = 87, normalized size = 0.71 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3 A \left (\cos ^{2}\left (d x +c \right )\right )+4 A \cos \left (d x +c \right )+8 A +15 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{15 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+8*A+15*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x
+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)

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maxima [B]  time = 0.66, size = 224, normalized size = 1.84 \[ \frac {\sqrt {2} {\left (30 \, \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 30 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 120 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/60*(sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/
5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/
2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 120*sqrt(2)*C*sqrt(a)*
sin(1/2*d*x + 1/2*c))/d

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mupad [B]  time = 3.86, size = 94, normalized size = 0.77 \[ \frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (35\,A\,\sin \left (c+d\,x\right )+60\,C\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )\right )}{30\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/(1/cos(c + d*x))^(5/2),x)

[Out]

(cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(35*A*sin(c + d*x) + 60*C*sin
(c + d*x) + 8*A*sin(2*c + 2*d*x) + 3*A*sin(3*c + 3*d*x)))/(30*d*(cos(c + d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/sec(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + C*sec(c + d*x)**2)/sec(c + d*x)**(5/2), x)

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